題目：
Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

Notice
Have you consider that the string might be empty? This is a good question to ask during an interview.

For the purpose of this problem, we define empty string as valid palindrome.

思路：
這題可以從左右指針來看，頭一個，尾一個，一路往中心走，走的同時要判斷是不是正確字元，可以用 isalnum 來解決，因為考慮大小寫，所以在比較的時候要使用 toupper (tolower 也行)，全部統一轉換成一種大小寫來比較即可，只要一出錯就 return false.

程式：

class Solution {
public:
    /**
     * @param s: A string
     * @return: Whether the string is a valid palindrome
     */
    bool isPalindrome(string &s) {
        // write your code here
        if (s.empty())
            return true;
        
        int l = 0;
        int r = s.size()-1;
        
        while( l < r ) {
            while(!isalnum(s[l]) && l < r)
                l++;
            while(!isalnum(s[r]) && l < r)
                r--;
            if (toupper(s[l]) != toupper(s[r]))
                return false;
            l++;
            r--;
        }
        return true;
    }
};